Question

Previously,

4​%

of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than

4​%

today. She randomly selects

125125

pregnant mothers and finds that

4

of them smoked 21 or more cigarettes during pregnancy. Test the​ researcher's statement at the

alpha equals 0.05α=0.05

level of significance.

What are the null and alternative​ hypotheses?

Upper H 0H0​:

muμ

equals=

nothing versus

Upper H 1H1​:

muμ

less than<

nothing

​(Type integers or decimals. Do not​ round.)

Because np 0 left parenthesis 1 minus p 0 right parenthesisnp01−p0equals=nothing

▼

​10, the normal model

▼

may not

may

be used to approximate the​ P-value.

​(Round to one decimal place as​ needed.)

Find the​ P-value.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

Is there sufficient evidence to support the​ obstetrician's statement?

A.

NoNo​,

rejectreject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There

is notis not

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4​%.

B.

YesYes​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There

isis

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4​%.

C.

NoNo​,

do not rejectdo not reject

the null hypothesis because the​ P-value is

greatergreater

than

alphaα.

There

is notis not

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

4​%.

D.

YesYes​,

rejectreject

the null hypothesis because the​ P-value is

lessless

than

alphaα.

There

isis

sufficient evidence to conclude that the percentage of mothers who smoke 21 or more cigarettes during pregnancy is less than

44%.

Answer 1

We are testing,

H0: p= 0.04 vs H1: p<0.04

Now, n= 125, np= 5, n(1-p) = 120

Since np<10, the normal may not be used to approximate the p-value.

However assuming it's Correct:

z statistic: -p/√p(1-p)/n

Now, = 4/125= 0.032

So test statistic: (0.032-0.04)/√0.04*0.96/125= -4.56

p-value: P(z<-4.56) <0.00001

Since the p-value <significance level of 0.05, we have sufficient evidence to Reject H0 at the 5% level of significance and conclude that yes p<0.04

So option D is correct.